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I need solve the follows limits:

$1)$ $\lim_{x\rightarrow 0}\frac{\sin x^2\cdot\sin x}{e^{5x^4}-1}$

$2)$ $\lim_{x\rightarrow 0}\frac{(e^{3x}-1)\sin 3x}{\ln(2x^2+1)}$

$3)$ $\lim_{x\rightarrow 0}\frac{x\ln(2x^2+1)}{\sin^3 x}$

$4)$ $\lim_{x\rightarrow 1}\frac{\sqrt[3]{2x-1}-1}{\sqrt[3]{2-x}-x}$

$5)$ $\lim_{n\rightarrow \infty} \left[\left(\frac{n^2-2n-3}{n^2+5n+5}\right)^\frac{n^2+1}{n}+\frac{5^n-3^n}{5^{n+1}-3^{n+1}}\right]$

My solutions:

$2)$ is $\frac{9}{2}$

$4)$ is $-\sqrt[3]{4}$

$5)$ is $\frac{1}{5}+e^{-7}$

but i didn`t now is it correct, and i know how to solve limits $1)$ and $3).$

Please help me. Thanks for your time.

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  • $\begingroup$ They are all rather easily solved with Taylor series. Are you allowed to use them? $\endgroup$
    – Alex
    Jan 6, 2016 at 9:09
  • $\begingroup$ I have not used the Taylor series because the examples are for high school $\endgroup$
    – user145717
    Jan 6, 2016 at 9:11
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    $\begingroup$ Why so many questions in a single Post? Use $$\lim_{x\to0}\dfrac{\sin x}x=1,\lim_{x\to0}\dfrac{e^x-1}x=1; \lim_{x\to0}\dfrac{\ln(1+x)}x=1$$ $\endgroup$ Jan 6, 2016 at 9:19
  • $\begingroup$ In my opinion 5) is OK. $\endgroup$
    – Leon
    Jan 6, 2016 at 9:57

2 Answers 2

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$2)$ $$\lim_{x\rightarrow 0}\frac{\frac{(e^{3x}-1)\sin 3x}{3x\cdot 3x}}{\frac{\ln(2x^2+1)}{9x^2}}=\lim_{x\to 0}\frac{\frac{e^{3x-1}}{3x}\cdot\frac{\sin3x}{3x}}{\frac{\ln(2x^2+1)}{9x^2}}= \lim_{x\rightarrow 0}\frac{1}{\frac{\ln(2x^2+1)}{9x^2}}=\lim_{x\rightarrow 0}\frac{1}{\frac{1}{9x^2}{\ln(2x^2+1)}}=\lim_{x\rightarrow 0}\frac{1}{{\ln(2x^2+1)}^\frac{1}{9x^2}}=\lim_{x\rightarrow 0}\frac{1}{{\ln(2x^2+1)}^{\frac{1}{2x^2}\cdot\frac{2}{9}}}=\frac{1}{\ln e^{\frac{2}{9}}}=\frac{9}{2}$$

$3)$ $$\lim_{x\rightarrow 0}\frac{x\ln(2x^2+1)}{\sin^3 x}=\lim_{x\rightarrow 0}\frac{\frac{x\ln(2x^2+1)}{x^3}}{\frac{\sin^3 x}{x^3}}=\lim_{x\rightarrow 0}\frac{\frac{x\ln(2x^2+1)}{x^3}}{\frac{\sin x}{x}\cdot\frac{\sin x}{x}\cdot\frac{\sin x}{x}}=\lim_{x\rightarrow 0}{\frac{\ln(2x^2+1)}{x^2}}$$ $$=\lim_{x\rightarrow 0}\frac{1}{x^2}\ln(2x^2+1)=\lim_{x\rightarrow 0}\frac{1}{x^2}\ln(2x^2+1)=\lim_{x\rightarrow 0}\ln(2x^2+1)^\frac{1}{x^2}=\lim_{x\rightarrow 0}\ln(2x^2+1)^{\frac{1}{2x^2}\cdot 2}=\ln e^2=2$$

$4)$ $$\lim_{x\rightarrow 1}\frac{\sqrt[3]{2x-1}-1}{\sqrt[3]{2-x}-x}=\lim_{x\rightarrow 1}\frac{\sqrt[3]{2x-1}-1}{\sqrt[3]{2-x}-x}\cdot\frac{\sqrt[3]{(2-x)^2}+x\sqrt[3]{2-x}+x^2}{\sqrt[3]{(2-x)^2}+x\sqrt[3]{2-x}+x^2}=\lim_{x\to 0}\frac{(\sqrt[3]{2x-1}-1)(\sqrt[3]{(2-x)^2}+x\sqrt[3]{2-x}+x^2)}{2-x-x}=\lim_{x\to 0}\frac{(\sqrt[3]{2x-1}-1)(\sqrt[3]{(2-x)^2}+x\sqrt[3]{2-x}+x^2)}{2-2x}=\lim_{x\to 0}\frac{(\sqrt[3]{2x-1}-1)(\sqrt[3]{(2-x)^2}+x\sqrt[3]{2-x}+x^2)}{2-2x}\cdot\frac{\sqrt[3]{(2x-1})^2+\sqrt[3]{2x-1}+1}{\sqrt[3]{(2x-1})^2+\sqrt[3]{2x-1}+1}=\lim_{x\to 0}\frac{(2x-1-1)\cdot(\sqrt[3]{(2-x)^2}+x\sqrt[3]{2-x}+x^2)}{(2-2x)(\sqrt[3]{(2x-1})^2+\sqrt[3]{2x-1}+1)}=-\sqrt[3] 4$$

$5)$ $$\lim_{n\rightarrow \infty} \left[\left(\frac{n^2-2n-3}{n^2+5n+5}\right)^\frac{n^2+1}{n}+\frac{5^n-3^n}{5^{n+1}-3^{n+1}}\right]=\lim_{n\to \infty}\left(\frac{n^2-2n-3}{n^2+5n+5}\right)^\frac{n^2+1}{n}+\lim_{n\to\infty}\frac{5^n-3^n}{5^{n+1}-3^{n+1}}=\lim_{n\to \infty}\left(\frac{n^2+5n+5-7n-8}{n^2+5n+5}\right)^\frac{n^2+1}{n}+\lim_{n\to\infty}\frac{5^n\left(1-\left(\frac{3}{5}\right)^n\right)}{5^n\left(5-3\cdot\left(\frac{3}{5}\right)^n\right)}$$ $$=\lim_{n\to \infty}\left(1+\frac{1}{\frac{n^2+5n+5}{-7n-8}}\right)^{{\frac{n^2+5n+5}{-7n-8}}\cdot{\frac{-7n-8}{n^2+5n+5}}\cdot{\frac{n^2+1}{n}}}+\frac{1}{5}=e^{-7}+\frac{1}{5}$$

We hope you have these solutions help

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All of these are trivial if you use taylor expansion; You have to know that around $0$,

$$\sin x \sim x$$ $$\ln(1 + x) \sim x$$ $$e^x \sim 1 + x$$ and so on. For example the third one, you can write

$$\frac{x \ln(2x^2 + 1)}{\sin^3 x} \sim \frac{x \cdot 2x^2}{x^3} = 2$$ so $$\lim_{x \to 0} \frac{x \ln(2x^2 + 1)}{\sin^3 x}= 2$$

try to do the rest yourself! If you're just looking for a comfirmation of your calculations, then just input those on wolfram alpha and it'll give you the answer

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  • $\begingroup$ It's unclear if the OP is allowed to use them $\endgroup$
    – Alex
    Jan 6, 2016 at 9:11
  • $\begingroup$ thanks sir, by $1)$ how to proced $\endgroup$
    – user145717
    Jan 6, 2016 at 9:21
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    $\begingroup$ For the first, since $\sin(x) \sim x$, $\sin(x^2)\sim x^2$, $e^y\sim 1+y$. Replace $y$ by $5x^4$. $\endgroup$ Jan 6, 2016 at 9:37

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